Question: Simplify; express your answer in exponential form. Assume $n\neq 0, t\neq 0$. $\dfrac{{(n^{-1})^{-2}}}{{(n^{-3}t^{-5})^{-5}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${n^{-1}}$ to the exponent ${-2}$ . Now ${-1 \times -2 = 2}$ , so ${(n^{-1})^{-2} = n^{2}}$ In the denominator, we can use the distributive property of exponents. ${(n^{-3}t^{-5})^{-5} = (n^{-3})^{-5}(t^{-5})^{-5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(n^{-1})^{-2}}}{{(n^{-3}t^{-5})^{-5}}} = \dfrac{{n^{2}}}{{n^{15}t^{25}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{2}}}{{n^{15}t^{25}}} = \dfrac{{n^{2}}}{{n^{15}}} \cdot \dfrac{{1}}{{t^{25}}} = n^{{2} - {15}} \cdot t^{- {25}} = n^{-13}t^{-25}$.